An oil film on top of water has one patch that is much thinner than the wavelength of visible light. The index of refraction of the oil is less than that of water. Will the reflection from that extremely thin part of the film be bright or dark? Explain.

they indicate that the thickness is much less than the length of anything we have a destructive interference, bone has a dark area

Explanation:

Let's analyze the interaction of the ray reflected on the surface of the oil and at the bottom of the film, to know what kind of interference we have.

- When the beam passes from the air to the oil film, it has a phase change of 180 since the refractive index of the oil is greater than the refractive index of the air n = 1

- When the beam reaches the oil-water interface, the refractive impact of the water is oil cobs, therefore it has a 180 phase change

- in the oil film because the speed of the beam is different from that of air and the frequency in the air and oil is the same, since the propagation of a resonant type process, to meet the equation

v = λ f

The wavelength must change c = λ₀ f

c / v = λ₀ / λ

n = λ₀ / λ

λₙ = λ₀ / n

In general this type of reflection is almost normal to the surface, so we can calculate the optical path difference, if t is the thickness of the film, for constructive interference

2t = m λₙ + λₙ / 2 + λₙ / 2

The last two terms are for the phase change

2t / = (m + 1) λₙ

This is the condition to see a bright beam

For the first reflection m = 0

2t = λₙ

Therefore the minimum thickness of the film for constructive interference is λₙ / 2.

As they indicate that the thickness is much less than the length of anything we have a destructive interference, bone has a dark area